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STRUCTURE OF SULFUR ISOTOPES
By Prof. L. Kaliambos (Natural philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Sulfur (S) has 25 known isotopes with mass numbers ranging from 26 to 49, four of which are stable: S-32 (95.02%), S-33 (0.75%), S-34(4.21%), and S-36 (0.02%). The preponderance of sulfur-32 is explained by its production from carbon-12 plus successive fusion capture of five helium nuclei, in the so-called alpha process of exploding type II supernovae (see silicon burning). Other than S-35, the radioactive isotopes of sulfur are all comparatively short-lived. S-35 is formed from cosmic ray spallation of Ar-40 in the atmosphere. It has a half-life of 87 days. The next longest-lived radioisotope is sulfur-38, with a half-life of 170 minutes. The shortest-lived is S-49, with a half-life shorter than 200 nanoseconds. WHY S-32, S-33, S-34 AND S-36 ARE STABLE NUCLIDES After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. For comparing the structure of S-32 with P-32 you can read my STRUCTURE OF P-32 AND S-32 . In the following diagram of S-32 you can see that the nucleons like the p13, n13, n14, p14, p15, n15, p16 and n16 make strong vertical rectangles with 6 strong vertical bonds and 8 weak horizontal bonds existing outside a parallelepiped with 6 horizontal planes of opposite spins ( core of Mg-24 with S=0). Note that the same nucleons as shown in the diagram of S-34 make two unstable rectangles outside the same core with S=0 having only 4 strong vertical bonds but many weak horizontal bonds (12 bonds) unable to overcome the nn and pp repulsions. However the S-34 with S=0 is a stable nuclide because the two blank positions formed by the protons of the two rectangles are able to receive the n17 and n18 at symmetrical positions which make two bonds per neutron. Here the 4 extra bonds increase the binding energies of weak bonds able to overcome the nn and pp repulsions. But in the unstable S-35 the only two bonds per neutron of the n17 cannot give enough energy to the weak horizontal bonds. So they cannot overcome the pp and nn repulsions. On the other hand the S-33 with S=+3/2 is also a stable nuclide. In the following diagram of S-33 you see that the core is a parallelepiped with 5 horizontal planes like the +HP1, -HP2, +HP3, -HP4, and +HP5 giving S =+2. In this case the 6 deuterons like the p11n11, p12n12, p13n13, p14n14, p15n15 and p16n16 make vertical rectangles with S=0 outside the core of 5 horizontal planes, and the n17 (-1/2)at -HP4 makes two bonds per neutron able to increase the binding energies of the horizontal bonds, which overcome the pp and nn repulsions. Since the core has S=+2 and the deuterons give S=0 we get the S-33 with S =+3/2 given by S = +2 + 0 + (-1/2) = +3/2. After a detailed study I found that the Sulfur isotopes are based on the structure of S-32, S-33 and S-34 . For example the stable S-36 is based on the S-34 by adding the extra n19 and n20 than those of S-34 which make also two bonds per neutron at symmetrical positions. ' DIAGRAM OF STABLE S-32 WITH S=0 ' ' . p12.......n12' ' -HP6 n11.......p11' ' . n10.......p10.......n16' ' +HP5 n15.....p9.......n9 ' ' . p8.........n8.......p14 ' ' -HP4 p13.......n7.......p7 ' ' . n6........p6.......n14' ' +HP3 n13.......p5.......n5 ' ' . p4........n4........p16' ' -HP2 p15......n3.........p3 ' ' . n2........p2' ' +HP1 p1........n1 ' DIAGRAM OF STABLE S-33 WITH S =+3/2 ' ' . n10......p10.....n16 ' +HP5 p9.......n9 .......p16 ' ' . n17.......p8........n8.......p15 ' ' -HP4 n7........ p7.......n15 ' ' . p12.......n6........p6.......n14' ' +HP3 n12.......p5........n5.......p14 ' ' . n11........p4........n4........p13' ' -HP2 p11......n3........p3.......n13 ' ' . n2........p2' ' +HP1 p1.......n1 ' ' ' ' ' ' DIAGRAM OF STABLE S-34 WITH S =0 '' '' ' . p12.......n12''' ' -HP6 n11.......p11' ' . n10.......p10' ' +HP5 n17.....p9.......n9 ' ' . n14.......p8.........n8.......p16 ' ' -HP4 p14.......n7.......p7........n16 ' ' . p13.......n6........p6.......n15' ' +HP3 n13.......p5.......n5.........p15 ' ' . p4........n4' ' -HP2 n3.........p3........n18 ' ' . n2........p2' ' +HP1 p1........n1 ' NUCLEAR STRUCTURE OF S-38, S-40, S-42, S-44, S-46 AND S-48 WITH S=0 The structure of the above unstable nuclides is based on the structure of P-34. Adding extra neutrons of opposite spins giving weak horizontal bonds in the structure of S-34 with S=0 we get the structure of the above unstable nuclides with the same S=0. For example the S-48 with S=0 has 14 more extra neutrons of opposite spins than those of S-34. Since the extra 14 neutrons give S=0 we get S=0. ' ' NUCLEAR STRUCTURE OFS-31 S-30, S-28 AND S-26 ' '''The structure of the S-31 with S =+1/2 is based on the structure of S=32 with S=0. In the absence of the n12(-1/2) which was at the corner of the parallelepiped we get the total S =+1/2 because S = 0 - (-1/2) = +1/2 . Moreover in the absence of n12(-1/2) and n1( +1/2) which were at the two corners of the parallelepiped we get the structure of S=30 with S=0. However the structure of S-28 with S=0 is based on the structure of S-34. In this case the two absent neutrons like the n13(+1/2) and n16((-1/2) from the corners of the rectangles along with the absent n1(+1/2) and n12(-1/2) from the corners of the parallelepiped contribute to the structure of S-28 with S=0. Also of in the absent of two more neutrons of opposite spins from the corners of the rectangles of S-28 we get the structure of S-26 with S=0. '''NUCLEAR STRUCTURE OF S-29 AND S-27 WITH S=+5/2' The structure of the above nuclides is based on the structure of S-33 with S=+3/2. For example in the absence of two neutrons of negative spins and of two neutrons of opposite spins we get the structure of S-29 with S=+5/2. That is S = +3/2 - 2(-1/2) - (+1/2) - (-1/2) = +5/2 Then following the structure of S-29 with S=+5/2 we see that in the absence of two more neutrons of opposite spins we get the structure of the unstable S-27 with the same S= +5/2. NUCL-EAR STRUCTURE OF S-35 AND S-39 WITH S=+3/2 Similarly the structure of the above nuclides is based on the structure of S-33 with S=+3/2 . For example adding two neutrons of opposite spins which make single horizontal bonds we get the structure of S-35 with the same S= +3/2. Also adding 6 extra neutrons of opposite spins in the structure of S=+3/2 we get the structure of S-39 with the same S=+3/2. ' NUCLEAR STRUCTURE OF S-45, S-47 AND S-49 WITH S=-3/2' In these cases the nucleons of the S-33 change the spins. For example we have -HP1, +HP2, -HP3, +HP4 and -HP5 giving S =-3/2. Thus adding 12 extra nucleons of opposite spins in the S-33 with S=-/32 we get the structure of S-45 with the same S=-3/2. Also adding two more extra neutrons of opposite spins in the structure of S-45 and 4 more neutrons of opposite spins we get the structures of S-47 and S-49 respectively with the same S=-3/2. ' NUCLEAR STRUCTURE OF S-37, S-41 AND S-43 WITH S =-7/2' Here following the structure of S-33 with S=-3/2 we see that when the deuteron p12n12 goes from the second horizontal plane to the first one we get a structure of S =-7/2. Thus adding 4 extra neutrons of opposite spins in this new arrangement of nucleons we get the structure of S-37 with S=-7/2. In the same way we get the structures of S-41 and S-43 by adding 8 extra neutrons or 10 extra neutrons of opposite spins to get the structures of S-41 and S-43 respectively. ' ' ' ' ' ' Category:Fundamental physics concepts